The equation for this sequence is:

a_n = a_n-1 + 2(n - 1)

where n is the term's order in the sequence, a_n is the value of the term, and a_n-1 is the value of the term before a_n. This makes the next number:

a₆ = a₅ + 2(6 - 1)
a₆ = 21 + 2(5)
a₆ = 31

The distributive property for division helps in solving expressions like \({b + c \over a}\). It specifies that the result of dividing a fraction with multiple terms in the numerator and one term in the denominator can be obtained by dividing each term individually and then totaling the results: \({b + c \over a} = {b \over a} + {c \over a}\). For example, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).

Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):

3 + (2 + 2) ÷ 5 x 5 - 2²
P: 3 + (4) ÷ 5 x 5 - 2²
E: 3 + 4 ÷ 5 x 5 - 4
MD: 3 + \( \frac{4}{5} \) x 5 - 4
MD: 3 + \( \frac{20}{5} \) - 4
AS: \( \frac{15}{5} \) + \( \frac{20}{5} \) - 4
AS: \( \frac{35}{5} \) - 4
AS: \( \frac{35 - 20}{5} \)
\( \frac{15}{5} \)
3

To add these radicals together their radicands must be the same:

2\( \sqrt{32} \) + 4\( \sqrt{2} \)
2\( \sqrt{16 \times 2} \) + 4\( \sqrt{2} \)
2\( \sqrt{4^2 \times 2} \) + 4\( \sqrt{2} \)
(2)(4)\( \sqrt{2} \) + 4\( \sqrt{2} \)
8\( \sqrt{2} \) + 4\( \sqrt{2} \)

Now that the radicands are identical, you can add them together:

To fill a 16 gallon tank exactly halfway you'll need 8 gallons of fuel. Each fuel can holds 2 gallons so:

cans = \( \frac{8 \text{ gallons}}{2 \text{ gallons}} \) = 4